Infinitesimal vs zero probabilities in uniform continuous distributions Suppose we select two numbers uniformly from $\mathbb{R}$: $x \in [0, a]$ and $y \in [0,b]$, where $a$ and $b$ are arbitrary nonnegative integers. 
I can visualize this as a 2-d rectangle with side lengths $a$ and $b$. Any randomly selected point will fall within or on the boundaries of the rectangle. This lets me answer a question like "What is $P(x < y)$?" by computing an area (in this case, the area under the main diagonal, which is $\frac{1}{2}$ .) 
Now if I was to try to answer "What is $P(x=y)$?" that is the main diagonal, and as a line, this has zero area, so $P(x=y)$ should be zero. 
Is this answer actually correct, or is my way of solving the problem wrong? I know that the probability of picking a specific point in a univariate continuous distribution is 0 and only intervals have nonzero probability, but is $P(x=y)$ exactly equal to 0 or is it some infinitesimally small nonzero quantity?
 A: You're correct. First off, if a probability is defined to be a real number in $[0,1]$, then a probability can't be an "infinitesimal but nonzero quantity," as such do not exist in $\Bbb{R}$ by the Archimean property. 
The formation of measure theory was partially motivated by problems in probability. Given a set $S\subset [0,a]\times[0,b]$, the probability that a uniformly chosen random variable lies in $S$ is equal to the Lebesgue measure of $S$ divided by $ab$. The Lebesgue measure is, essentially, the best formal definition for the area/volume/etc. of a subset of $\Bbb{R}^n$ (well, not all subsets; there are some "unmeasureable subsets" of $\Bbb{R}^n$). The Lebesgue measure of a line in $\Bbb{R}^2$ is zero. 
A: Even though the diagonal contains uncountably many points, it still has measure $0$, and therefore the probability of a point being on it is $0$. It may be that this would be different in non-standard analysis, where numbers exist that are larger than $0$ but smaller than any positive real.
A: Here is one way of looking at it.
Consider the $n$ squares $[{k \over n} ,({k \over n}+{1 \over n }) a] \times [{k \over n}, ({k \over n}+{1 \over n}) b]$, with $k=0,...,n-1$.
Then you can see that $\Delta=\{(x,y) | x=y \}$ is contained in these squares fro any $n$, and the
total volume of the squares is ${1 \over n} ab$.
Hence the volume of $\Delta$ is zero.
